Find $\int \sec^3 \theta d\theta$.

$\frac{1}{2}[ \sec \theta \tan \theta + \ln|\sec \theta + \tan \theta| + C]$

The correct answer is Option (2) → $\frac{1}{2}[ \sec \theta \tan \theta + \ln|\sec \theta + \tan \theta| + C]$

Let $I = \int \sec^3 \theta d\theta = \int \sec \theta \sec^2 \theta d\theta$ ... (i)

Integrating by parts, we have

$u = \sec \theta, v = \sec^2 \theta$

So, $I = \int u v d\theta$

$I = u \int v d\theta - \int \left( \frac{du}{d\theta} \int v d\theta \right) d\theta$

$I = \sec \theta \int \sec^2 \theta d\theta - \int \left\{ \frac{d(\sec \theta)}{d\theta} \cdot \int \sec^2 \theta d\theta \right\} d\theta$

$I = \sec \theta \tan \theta - \int \sec \theta \tan \theta \cdot \tan \theta d\theta$

$I = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d\theta$

$I = \sec \theta \tan \theta - \int \sec \theta (\sec^2 \theta - 1) d\theta$

$I = \sec \theta \tan \theta - \int \sec^3 \theta d\theta + \int \sec \theta d\theta$

$I = \sec \theta \tan \theta - I + \ln |\sec \theta + \tan \theta| + C$

$2I = \sec \theta \tan \theta + \ln |\sec \theta + \tan \theta| + C$

$I = \frac{1}{2} [\sec \theta \tan \theta + \ln |\sec \theta + \tan \theta| + C]$